RESULTANT PRESSURE 



27 



In this case the resultant pressure will, for equilibrium, be vertical and equal to the weight 

 of the water ; 



.. Resultant pressure = 64-2 x ^v r 3 Ibs. 



In the case of a plane surface, such as a dock gate,- the total pressure 

 on a face will be the same as the resultant pressure, both being normal to 

 the face. 



Example 2. In the dock gate shown in Fig. 11, the width of gate being 20 feet, the depth 

 of the centroid of the submerged portion on the right-hand side is 5 feet, and on the left-hand 

 side is 2 feet. The areas of these submerged surfaces are 200 feet and 80 feet respectively, so 

 that the total pressures, and also the resultant pressures on the two faces, are 5 x 200 x 62-4 = 

 62,400 Ibs. and 2 x 80 x 62-4 = 

 10,000 Ibs. The resultant of the 

 two pressures will then be a single 

 force of 62,400 - 10,000 = 52,400 

 Ibs. acting from right to left. The 

 magnitude of the resultant pressure 

 intensity, and of its distribution 

 over the gate are indicated in Fig. 

 11. Here the pressure intensity at 

 any depth is indicated by the hori- 

 zontal distance between the surface 

 of the gate and the straight lines 

 A C and F 6. The resultant pres- 

 sure at any depth is then to the left, 

 and is represented by the hori- 

 zontal width of the shaded area. 

 Evidently at all points below the 

 lower surface level, the resultant 

 pressure intensity will be constant, 

 since the pressure intensity in- 

 creases at equal rates on both sides 

 of the gate. The resultant force 

 to the left per foot run of the gate 

 is represented by the shaded area. 



The determination of the 

 position of the Centre of Pressure of an area is the same as that of the 

 line of action of the resultant of a series of statical forces each normal to 

 the surface under consideration. Referring this to rectangular co-ordi- 

 nates X and Y (Fig. 12) of which X is vertical and O Y in the 

 surface, we have, taking moments about these axes and representing the 

 co-ordinates of the Centre of Pressure by X Y, 



2 (p & A cos 0) X = 2 (p B A cos 6 x) } (1) 



2 (p B A cos 0) Y = 2 (p B A cos y) j (2) 



writing p = W x these become 



2 (W 8 A cos x) X = 2 (W B A cos x 2 ) ) (3) 



2 (W B A cos Ox) Y = 2(W8A cos0xy) j (4) 



